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Battery powered Current sink with low voltage cutoff
- Thread starter gregfox
- Start date
KrisBlueNZ
Sadly passed away in 2015
U2 regulates the +8V reference voltage. The 33k/15k voltage divider feeding U2 pin 1 sets the voltage (see the TL431 data sheet).
The other resistors set the voltages at both ends of the two potentiometers (VRT and VRC). VRT needs 8V at the top end and 4V at the bottom end, to set the limits for the switch-off voltage range. VRC needs 1.5V at the top end and 0V at the bottom end, to give a current sink range of 0~1.5A using a 1 ohm shunt resistance.
The trimpots are used to set these voltages exactly. They are needed because the end-to-end resistance of a potentiometer typically is not very accurately controlled; tolerances of 20% and 10% are common.
The reason why I put a 33k resistor in series with VRM is to make the VRM adjustment more accurate. Assuming VRC has an end-to-end resistance tolerance of +/- 20%, the combined resistance of the 33k resistor and VRM needs to be adjustable between roughly 34.7k and 52k. If I had omitted the resistor, I would have had to specify a 100k trimpot for VRM. This would work, but the useful part of its range would have been from 34.7% to 52% (only 17% of its full adjustment range), so it would have been much more difficult to set accurately. The 33k resistor is called a "stopper resistor"; it narrows the adjustment range for the trimpot so you can set it more accurately.
Re increasing the 8V reference voltage to 9V, you can do this by changing the voltage divider that feeds U2 pin 1, but I wouldn't increase it much beyond that. The current through U2 is equal to (11V_rail_voltage - reference_voltage) / RR. For proper operation of U2, it must be at least 1 mA. The closer the reference voltage is to the 11V rail voltage, the more the first part of that formula, 11V_rail_voltage - reference_voltage, becomes dependent on the 11V rail voltage, which varies according to the amount of charge in the 11.1V battery. So the more you increase the 8V reference voltage, the more the current through U2 will be dependent on the 11.1V battery voltage.
If you can't PM me the link to the new thread about your battery-powered power supply, you can post the link on this thread.
Dave Jones of eevblog.com has done some work on low-power power supplies. He designed a nice little isolated USB-powered box. You might want to look at his designs for some ideas.
The other resistors set the voltages at both ends of the two potentiometers (VRT and VRC). VRT needs 8V at the top end and 4V at the bottom end, to set the limits for the switch-off voltage range. VRC needs 1.5V at the top end and 0V at the bottom end, to give a current sink range of 0~1.5A using a 1 ohm shunt resistance.
The trimpots are used to set these voltages exactly. They are needed because the end-to-end resistance of a potentiometer typically is not very accurately controlled; tolerances of 20% and 10% are common.
The reason why I put a 33k resistor in series with VRM is to make the VRM adjustment more accurate. Assuming VRC has an end-to-end resistance tolerance of +/- 20%, the combined resistance of the 33k resistor and VRM needs to be adjustable between roughly 34.7k and 52k. If I had omitted the resistor, I would have had to specify a 100k trimpot for VRM. This would work, but the useful part of its range would have been from 34.7% to 52% (only 17% of its full adjustment range), so it would have been much more difficult to set accurately. The 33k resistor is called a "stopper resistor"; it narrows the adjustment range for the trimpot so you can set it more accurately.
Re increasing the 8V reference voltage to 9V, you can do this by changing the voltage divider that feeds U2 pin 1, but I wouldn't increase it much beyond that. The current through U2 is equal to (11V_rail_voltage - reference_voltage) / RR. For proper operation of U2, it must be at least 1 mA. The closer the reference voltage is to the 11V rail voltage, the more the first part of that formula, 11V_rail_voltage - reference_voltage, becomes dependent on the 11V rail voltage, which varies according to the amount of charge in the 11.1V battery. So the more you increase the 8V reference voltage, the more the current through U2 will be dependent on the 11.1V battery voltage.
If you can't PM me the link to the new thread about your battery-powered power supply, you can post the link on this thread.
Dave Jones of eevblog.com has done some work on low-power power supplies. He designed a nice little isolated USB-powered box. You might want to look at his designs for some ideas.
U2
Dave Jones is awesome, I wish I knew about him before. Thank you for the explanation about the parts, as always very complete and easy to understand.
I will use the formula Vout = Vref(1+R1/R2) if I change the voltage on u2. I ordered the AP431A part which is 200mA, and that was the cause of my confusion. I’ll upload the Battery power supply as soon as I make a readable schematic.
Dave Jones is awesome, I wish I knew about him before. Thank you for the explanation about the parts, as always very complete and easy to understand.
I will use the formula Vout = Vref(1+R1/R2) if I change the voltage on u2. I ordered the AP431A part which is 200mA, and that was the cause of my confusion. I’ll upload the Battery power supply as soon as I make a readable schematic.
KrisBlueNZ
Sadly passed away in 2015
Right. My design also requires a voltage source to remain ON. If the switch is held at the ON position, the circuit will stay ON, but if there's no voltage source connected, as soon as the switch is returned to the middle position, it will turn off.
I don't see how this is a problem. What's the use of having the circuit switched on, if there's no voltage source connected to it?
Other comments on that schematic:
1. Q3's source and drain are exchanged
2. You only need one trimpot to set the maximum current; you can delete the second trimmer and R3. Then you don't need IC2A either.
3. You should protect IC1D against input voltages greater than VCC. When the voltage source is initially connected, the circuit may not be powered up.
4. Your cells are still upside down. The polarity markings are right but the symbols are reversed.
Re your idea of using a microcontroller instead of potentiometers (from your PM), sure you can do that, but consider the resolution of the DAC. An 8-bit DAC only gives you 256 output values, so you'll only be able to set your current with a resolution of 6 mA. If that's enough, then great.
Re using two meters instead of one, whether that's important depends on how you're going to use it. You could use two separate meters, and switch the voltage meter between the switch-off voltage and the actual voltage using a SPDT pushbutton. Hold the button in and adjust the turn-off voltage, then release the button to watch the terminal voltage. Then you could watch the voltage and current simultaneously. That might be useful when you're discharging a battery. What do you think?
I don't see how this is a problem. What's the use of having the circuit switched on, if there's no voltage source connected to it?
Other comments on that schematic:
1. Q3's source and drain are exchanged
2. You only need one trimpot to set the maximum current; you can delete the second trimmer and R3. Then you don't need IC2A either.
3. You should protect IC1D against input voltages greater than VCC. When the voltage source is initially connected, the circuit may not be powered up.
4. Your cells are still upside down. The polarity markings are right but the symbols are reversed.
Re your idea of using a microcontroller instead of potentiometers (from your PM), sure you can do that, but consider the resolution of the DAC. An 8-bit DAC only gives you 256 output values, so you'll only be able to set your current with a resolution of 6 mA. If that's enough, then great.
Re using two meters instead of one, whether that's important depends on how you're going to use it. You could use two separate meters, and switch the voltage meter between the switch-off voltage and the actual voltage using a SPDT pushbutton. Hold the button in and adjust the turn-off voltage, then release the button to watch the terminal voltage. Then you could watch the voltage and current simultaneously. That might be useful when you're discharging a battery. What do you think?
Last edited:
KrisBlueNZ
Sadly passed away in 2015
Re point 3, D1 is a good idea but it doesn't protect IC1D from having voltage on its input when there is no power to the circuit. This is the function of RP and DV in my design.
Thanks Kris, Didn't think of that, D2 and R3 should take care of the protection problem. I don't know if the design will fly as the scope shows about 30mV of jitter as the cutoff voltage point is approached by about .4V, and turns off at about .25V prior to reaching the cutoff point. Scratching head on this one.
Regards, Greg
Regards, Greg
Attachments
Replaced MTP3055
Replaced MTP3055 with BDX53. Circuit a bit more stable, but still has a problem.
At higher current flow, as the cutoff voltage is approached, the voltage at pin 12 on IC1D starts to sag, and pin 13 displays a 10 to 50mA "jitter"
As a result pin 14 of IC1D will go high before the cutoff voltage is reached, thus turning off the whole device.
G.
Replaced MTP3055 with BDX53. Circuit a bit more stable, but still has a problem.
At higher current flow, as the cutoff voltage is approached, the voltage at pin 12 on IC1D starts to sag, and pin 13 displays a 10 to 50mA "jitter"
As a result pin 14 of IC1D will go high before the cutoff voltage is reached, thus turning off the whole device.
G.
Attachments
KrisBlueNZ
Sadly passed away in 2015
D2 doesn't need to be a 3A diode. A 1N914 or 1N4148 will be fine.
Re the problem at high current. This may be caused by imperfect connections in the breadboard layout. Make the high-current path external to the breadboard, using all soldered connections. That means the connection from the power supply under test to the BDX53, including the emitter shunt resistor. That whole loop needs to be external to the breadboard. Run wires from the bottom of the emitter shunt, and from the collector, to the breadboard, to connect to the rest of the circuit.
Re the problem at high current. This may be caused by imperfect connections in the breadboard layout. Make the high-current path external to the breadboard, using all soldered connections. That means the connection from the power supply under test to the BDX53, including the emitter shunt resistor. That whole loop needs to be external to the breadboard. Run wires from the bottom of the emitter shunt, and from the collector, to the breadboard, to connect to the rest of the circuit.
Well, I soldered all the connections that you mentioned but still had now, 300mV of jitter. Imagine a scope on AC and 100 mV all green. When I touched some parts of the bread board it reduced to 50mV, leading me to believe you were right. I must have imperfect connections in the breadboard. I guess the only way to prove the circuit is to build it up with all colder connections. I’ll do that in a day or two, and report back. Again many thanks.
KrisBlueNZ
Sadly passed away in 2015
I've got a few suggestions and questions.
1. Is the load circuit powered from a battery as designed?
2. Is there any other connection from the load circuit to mains earth, or to the power supply you are loading in the test?
3. Can you see any waveform in the noise? Try the scope at different speeds and adjust the trigger threshold so it's within the noise.
4. R3 and D2 are in the wrong place; they need to be between the collector of the current sink transistor and the input of the comparator, not between the collector and the power supply under test.
5. I assume that error is only present on the schematic, not on your prototype. Are there any other differences between the prototype and the schematic?
6. I would remove C2 (across the power supply under test). If you want to keep it in the circuit, at least test with it removed and see whether it makes any difference.
7. Is the 11V rail clean?
8. Is there any AC signal on the base of the current sink transistor?
9. Have you connected a decoupling capacitor across the positive and negative power supply pins of the op-amp IC?
10. The op-amp is shown as two ICs, IC1 and IC2. I assume this is just another difference between the prototype and the schematic.
11. While you're at it, you might want to rotate Q3 clockwise 90 degrees. This just makes the schematic a bit clearer.
1. Is the load circuit powered from a battery as designed?
2. Is there any other connection from the load circuit to mains earth, or to the power supply you are loading in the test?
3. Can you see any waveform in the noise? Try the scope at different speeds and adjust the trigger threshold so it's within the noise.
4. R3 and D2 are in the wrong place; they need to be between the collector of the current sink transistor and the input of the comparator, not between the collector and the power supply under test.
5. I assume that error is only present on the schematic, not on your prototype. Are there any other differences between the prototype and the schematic?
6. I would remove C2 (across the power supply under test). If you want to keep it in the circuit, at least test with it removed and see whether it makes any difference.
7. Is the 11V rail clean?
8. Is there any AC signal on the base of the current sink transistor?
9. Have you connected a decoupling capacitor across the positive and negative power supply pins of the op-amp IC?
10. The op-amp is shown as two ICs, IC1 and IC2. I assume this is just another difference between the prototype and the schematic.
11. While you're at it, you might want to rotate Q3 clockwise 90 degrees. This just makes the schematic a bit clearer.
1. Is the load circuit powered from a battery as designed?
Answer: I tried that and it made no difference.
2. Is there any other connection from the load circuit to mains earth, or to the power supply you are loading in the test?
Answer: I’m using a triple isolated output power supply, and I don’t believe there is load to earth.
3. Can you see any waveform in the noise? Try the scope at different speeds and adjust the trigger threshold so it's within the noise.
Answer: Yes it’s a very nice sine wave, about 2uS See attached pix.
4. R3 and D2 are in the wrong place; they need to be between the collector of the current sink transistor and the input of the comparator, not between the collector and the power supply under test.
Answer: I think I fixed that see attached Schematic.
5. I assume that error is only present on the schematic, not on your prototype. Are there any other differences between the prototype and the schematic?
Answer: I hope not.
6. I would remove C2 (across the power supply under test). If you want to keep it in the circuit, at least test with it removed and see whether it makes any difference.
Answer: no difference.
7. Is the 11V rail clean?
Answer: Could be cleaner, I’ll test it with just batteries.
8. Is there any AC signal on the base of the current sink transistor?
Answer: No.
9. Have you connected a decoupling capacitor across the positive and negative power supply pins of the op-amp IC?
Answer: Yes .01uF
10. The op-amp is shown as two ICs, IC1 and IC2. I assume this is just another difference between the prototype and the schematic.
Answer: Yes, sorry I’ll fix that later
11. While you're at it, you might want to rotate Q3 clockwise 90 degrees. This just makes the schematic a bit clearer.
Answer
one
THANKS!
Answer: I tried that and it made no difference.
2. Is there any other connection from the load circuit to mains earth, or to the power supply you are loading in the test?
Answer: I’m using a triple isolated output power supply, and I don’t believe there is load to earth.
3. Can you see any waveform in the noise? Try the scope at different speeds and adjust the trigger threshold so it's within the noise.
Answer: Yes it’s a very nice sine wave, about 2uS See attached pix.
4. R3 and D2 are in the wrong place; they need to be between the collector of the current sink transistor and the input of the comparator, not between the collector and the power supply under test.
Answer: I think I fixed that see attached Schematic.
5. I assume that error is only present on the schematic, not on your prototype. Are there any other differences between the prototype and the schematic?
Answer: I hope not.
6. I would remove C2 (across the power supply under test). If you want to keep it in the circuit, at least test with it removed and see whether it makes any difference.
Answer: no difference.
7. Is the 11V rail clean?
Answer: Could be cleaner, I’ll test it with just batteries.
8. Is there any AC signal on the base of the current sink transistor?
Answer: No.
9. Have you connected a decoupling capacitor across the positive and negative power supply pins of the op-amp IC?
Answer: Yes .01uF
10. The op-amp is shown as two ICs, IC1 and IC2. I assume this is just another difference between the prototype and the schematic.
Answer: Yes, sorry I’ll fix that later
11. While you're at it, you might want to rotate Q3 clockwise 90 degrees. This just makes the schematic a bit clearer.
Answer
oneTHANKS!
Attachments
KrisBlueNZ
Sadly passed away in 2015
Now you've put the 10k resistor in series with the current sink transistor! It needs to go in series with the connection into the comparator, and the 1N914 should be on the left side of it (i.e. also connected to the comparator input). Like on my schematic in post #14.
So something's oscillating at 500 kHz... If there's no AC at the base of the current sink transistor, that seems to indicate that it's the power supply under test that's oscillating. Have you tried loading something else? Like a charged battery?
You might want to upload a photo of your prototype setup.
So something's oscillating at 500 kHz... If there's no AC at the base of the current sink transistor, that seems to indicate that it's the power supply under test that's oscillating. Have you tried loading something else? Like a charged battery?
You might want to upload a photo of your prototype setup.
Moved R7 & D3
As you can see, I really stink at protection circuits. Too late on the breadboard, I tore it down and started to solder it up on a pref-board. I should have it ready soon and will then try a battery in place of the power supply for both power and sink.
By the way I love the way the resistor/trimmer work, for the set current drain. Please correct me if I'm wrong, but by using the BDX53C rated at 8A could I not go up to 2A on my circuit if I had a large enough heat sink?
Many thanks,
G.
As you can see, I really stink at protection circuits. Too late on the breadboard, I tore it down and started to solder it up on a pref-board. I should have it ready soon and will then try a battery in place of the power supply for both power and sink.
By the way I love the way the resistor/trimmer work, for the set current drain. Please correct me if I'm wrong, but by using the BDX53C rated at 8A could I not go up to 2A on my circuit if I had a large enough heat sink?
Many thanks,
G.
Attachments
KrisBlueNZ
Sadly passed away in 2015
Yes, you can go up to 2A. The limit is the power dissipation in the transistor. You mustn't exceed its rated dissipation, which is 60W. There is also a safe operating area limit, shown in the data sheet as a graph of collector voltage versus collector current. According to the SOAR graph in the Fairchild data sheet, for collector voltages up to 20V the collector current may be up to 3A without going outside the safe operating area.
And of course the power dissipated is converted into heat and you need to keep the junction temperature under 150 degrees Celsius (preferably well under 100 degrees C to avoid burning yourself) so you need a heatsink with a sufficiently low thermal resistance, which is specified in degrees Celsius per watt dissipated, and is relative to the ambient temperature. For example if the transistor has 10V across it (collector-emitter) and there is 2A flowing through it (collector-emitter) it will dissipate 20 watts. With a heatsink with a free air (i.e. ventilated) thermal resistance of 4 degrees C per watt, the heatsink will rise by 80 degrees C above the ambient temperature. You can use fan cooling to improve the heatsink's thermal resistance.
And of course the power dissipated is converted into heat and you need to keep the junction temperature under 150 degrees Celsius (preferably well under 100 degrees C to avoid burning yourself) so you need a heatsink with a sufficiently low thermal resistance, which is specified in degrees Celsius per watt dissipated, and is relative to the ambient temperature. For example if the transistor has 10V across it (collector-emitter) and there is 2A flowing through it (collector-emitter) it will dissipate 20 watts. With a heatsink with a free air (i.e. ventilated) thermal resistance of 4 degrees C per watt, the heatsink will rise by 80 degrees C above the ambient temperature. You can use fan cooling to improve the heatsink's thermal resistance.
Schematic 7B
, It took me much longer to make the board then I thought. The big problem was, I had a 1k resister where I should have had a 10k. It took forever to find the mistake. I’ve attached a new Schematic and a trace. I found that you were right there is AC on the base of T1 (BDX53) The trace I took (S8000741) was with the unit under battery power, with a battery as the PSU in test. I’m sorry the pix is so bad, guess I have to upgrade my camera. The divisions are 50mV, so as you can see it’s about 300mV P to P.
This oscillation occurs at various times;
If sinking 10 mA it occurs at about .02-.04 volts before the cutoff voltage.
If sinking 1500 mA it occurs at about .25-.35 volts before the cutoff voltage.
All other time it looks like a straight line. Of course when this oscillation starts IC1A-1 goes high and shuts the device off.
I put in two meters instead of one. A 4 ½ digit for the cutoff voltage and a 4 digit for the current (not shown on the schematic). Other then the oscillation it works great.
Can you figure out why it does what it does, and any thoughts how to remedy the situation?
, It took me much longer to make the board then I thought. The big problem was, I had a 1k resister where I should have had a 10k. It took forever to find the mistake. I’ve attached a new Schematic and a trace. I found that you were right there is AC on the base of T1 (BDX53) The trace I took (S8000741) was with the unit under battery power, with a battery as the PSU in test. I’m sorry the pix is so bad, guess I have to upgrade my camera. The divisions are 50mV, so as you can see it’s about 300mV P to P.
This oscillation occurs at various times;
If sinking 10 mA it occurs at about .02-.04 volts before the cutoff voltage.
If sinking 1500 mA it occurs at about .25-.35 volts before the cutoff voltage.
All other time it looks like a straight line. Of course when this oscillation starts IC1A-1 goes high and shuts the device off.
I put in two meters instead of one. A 4 ½ digit for the cutoff voltage and a 4 digit for the current (not shown on the schematic). Other then the oscillation it works great.
Can you figure out why it does what it does, and any thoughts how to remedy the situation?
Attachments
KrisBlueNZ
Sadly passed away in 2015
First, there are three unrelated issues with your latest schematic.
1. You seem to have accidentally changed D1 from a high-current diode (1N5404 etc) to a 1N914. D3 should be a 1N914 but D1 still needs to be a high-current diode.
2. This is a drafting issue. Back in the day of printed and photocopied schematic diagrams, schematics were drawn so that four wires would never connect together at a single junction. Whenever four wires met at a single point, they were actually two wires crossing over each other. Following this convention meant that if a dot happened to appear at the intersection two wires where they cross each other, due to dirt or some other marking on the paper or during photocopying, it would be detected, and wouldn't cause the diagram to be misinterpreted. Some of us older EEs regard the connections at both ends of R15 as a mistake.
3. Also a drafting issue. The symbol you've used for Q3 doesn't include the reverse-connected diode between source and drain (called the "body diode"), although it is present in the MPF990. I find it's a good idea to include it on the schematic as a reminder that it's there. (I caught myself out once, before I started including the diode in the symbol.)
When the circuit is oscillating, does the 11V rail have AC on it as well? I think it will. I think the problem is simply that as soon as the threshold is reached, the circuit starts to shut down, and this affects the circuitry that detects the threshold being reached, causing it to start up, and so on, ad infinitum.
This is why I added the complicated logic to the shutdown circuit in my design. I've added a small capacitor (CH on the updated schematic below) to make my design even more resistant to this problem. You need to modify your design so that as soon as the cutoff point is detected, the decision is made to shut down. Then as the circuit shuts down, the decision that has been made is not affected. Or you could just copy the rest of my design.
1. You seem to have accidentally changed D1 from a high-current diode (1N5404 etc) to a 1N914. D3 should be a 1N914 but D1 still needs to be a high-current diode.
2. This is a drafting issue. Back in the day of printed and photocopied schematic diagrams, schematics were drawn so that four wires would never connect together at a single junction. Whenever four wires met at a single point, they were actually two wires crossing over each other. Following this convention meant that if a dot happened to appear at the intersection two wires where they cross each other, due to dirt or some other marking on the paper or during photocopying, it would be detected, and wouldn't cause the diagram to be misinterpreted. Some of us older EEs regard the connections at both ends of R15 as a mistake.
3. Also a drafting issue. The symbol you've used for Q3 doesn't include the reverse-connected diode between source and drain (called the "body diode"), although it is present in the MPF990. I find it's a good idea to include it on the schematic as a reminder that it's there. (I caught myself out once, before I started including the diode in the symbol.)
When the circuit is oscillating, does the 11V rail have AC on it as well? I think it will. I think the problem is simply that as soon as the threshold is reached, the circuit starts to shut down, and this affects the circuitry that detects the threshold being reached, causing it to start up, and so on, ad infinitum.
This is why I added the complicated logic to the shutdown circuit in my design. I've added a small capacitor (CH on the updated schematic below) to make my design even more resistant to this problem. You need to modify your design so that as soon as the cutoff point is detected, the decision is made to shut down. Then as the circuit shuts down, the decision that has been made is not affected. Or you could just copy the rest of my design.
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